Kenneth ChengMathematics · models · ideas
Search中文

PAPER · REASONING · PRACTICE

Rational expressions: meaning, domain and cancellation

Separate zero numerators from zero denominators and retain restrictions after cancellation.

40 minutes · Year 8, upper volume · text pp. 79–82 · foundation lessons 40–41

Textbook reference: Year 8, upper · Chapter 15: rational expressions · lessons 40–41

Chapter core practice: use after the referenced textbook lessons, or for chapter review. See the upper-volume plan

One worked example, one guided attempt, eight core questions, four follow-ups and two extensions. Write on paper; the teacher controls the reveal.

Teacher referenceArrange in My Classroom

  1. 4′Check prerequisites
  2. 5′Observe and predict
  3. 8′Worked example and guided attempt
  4. 15′Independent practice
  5. 5′Discuss and correct
  6. 3′Exit questions

Observe · explain

A question before practising

Compare (x²−9)/(x−3) and x+3 at several allowed inputs, then inspect x=3. Why is the restriction still needed?

Organise the key idea

A rational expression is defined only when its denominator is nonzero. Cancel factors, not terms; cancelling a factor never restores a forbidden input.

Open the original chapter

Use the lesson question here for paper discussion; the original activity opens separately.

Worked example1

Simplify and retain every restriction from the original expression.

x2−9x−3\frac{x^2-9}{x-3}
Check your work
  • Use factor cancellation or a correct common denominator.
  • Retain all original restrictions, including a nonzero divisor.
Guided attempt2

Simplify and retain every restriction from the original expression.

2x2+4x2x\frac{2x^2+4x}{2x}
  1. Original restriction: x≠___.
  2. Factor the numerator as 2x(___); cancel the factor ___, not added terms.
Check your work
  • Use factor cancellation or a correct common denominator.
  • Retain all original restrictions, including a nonzero divisor.
Prerequisite3

Which displayed expression has a variable in its denominator?

1x+2\frac{1}{x+2}
  1. x²
  2. x+1
  3. 1/(x+2)
  4. x/3
Check your work
  • Choose the statement justified by the stated conditions and explain why.
Prerequisite4

Evaluate 1/(x−2) at x=3. Check the denominator first.

1/(3−2)1/(3-2)
Check your work
  • Use the stated relationship and show the calculation.
  • State the result with its unit where appropriate.
Independent practice5

State the domain restriction. Does the fraction simplify by cancellation?

x+1x−4\frac{x+1}{x-4}
Check your work
  • Use factor cancellation or a correct common denominator.
  • Retain all original restrictions, including a nonzero divisor.
Independent practice6

For which x is (x−5)/(x+1) equal to zero? Check that the expression is defined.

x−5x+1=0\frac{x-5}{x+1}=0
Check your work
  • Show a complete equation after clearing denominators.
  • Check original denominators and the original equality.
Independent practice7

Simplify and retain every restriction from the original expression.

x2−16x+4\frac{x^2-16}{x+4}
Check your work
  • Use factor cancellation or a correct common denominator.
  • Retain all original restrictions, including a nonzero divisor.
Independent practice8

Simplify and retain every restriction from the original expression.

x2+3xx\frac{x^2+3x}{x}
Check your work
  • Use factor cancellation or a correct common denominator.
  • Retain all original restrictions, including a nonzero divisor.
Exit question9

Simplify (x−2)/(x−2). Does the simplified value permit x=2? State the original restriction.

x−2x−2\frac{x-2}{x-2}
Check your work
  • Use factor cancellation or a correct common denominator.
  • Retain all original restrictions, including a nonzero divisor.
Exit question10

Simplify and retain every restriction from the original expression.

x+2x\frac{x+2}{x}
Check your work
  • Use factor cancellation or a correct common denominator.
  • Retain all original restrictions, including a nonzero divisor.
Further practice11

Simplify and retain every restriction from the original expression.

3x6x\frac{3x}{6x}
Check your work
  • Use factor cancellation or a correct common denominator.
  • Retain all original restrictions, including a nonzero divisor.
Further practice12

Simplify and retain every restriction from the original expression.

x2−1x−1\frac{x^2-1}{x-1}
Check your work
  • Use factor cancellation or a correct common denominator.
  • Retain all original restrictions, including a nonzero divisor.
Further practice13

Simplify and retain every restriction from the original expression.

x2+2xx+2\frac{x^2+2x}{x+2}
Check your work
  • Use factor cancellation or a correct common denominator.
  • Retain all original restrictions, including a nonzero divisor.
Further practice14

Simplify and retain every restriction from the original expression.

x−1x+1\frac{x-1}{x+1}
Check your work
  • Use factor cancellation or a correct common denominator.
  • Retain all original restrictions, including a nonzero divisor.
Extension15

After cancellation the formula is x+2. Explain why it is not the same original expression on ALL real inputs.

x2−4x−2\frac{x^2-4}{x-2}
Check your work
  • Use factor cancellation or a correct common denominator.
  • Retain all original restrictions, including a nonzero divisor.
Extension16

Simplify, and explain why x=1 remains forbidden even though the final formula is just x.

x2−xx−1\frac{x^2-x}{x-1}
Check your work
  • Use factor cancellation or a correct common denominator.
  • Retain all original restrictions, including a nonzero divisor.
Preparation and assessment notes

Prerequisites

Fractions, linear equations and factorisation.

Look for in students’ work

Use the worked model and guided attempt selectively, then the eight core questions. Keep further practice and extensions optional. Ask for a reason and an original-equation or expansion check.

Collect the paper exit questions and note completion and common errors. This page does not grade or store paper answers.

Paper for the class

Download all worksheets (ZIP · 318 PDFs · 47.2 MB)

All 53 Year 7–8 upper- and lower-volume packs in Traditional Chinese and English. Includes teacher answers; student core, further practice and teacher papers are in separate folders.

Core student worksheet: two A4 pages. Further practice is separate. Teacher copies include reference reasoning.

Textbook reference and authorship

Chapter-core practice after the mapped textbook lessons; these tasks do not replace every first-teaching period.

大湾区初中数学三维课堂（八年级上册）正文.pdf

Year 8, upper volume · text pp. 79–82 · foundation lessons 40–41

Original questions and explanations; the source textbook and answer PDFs are not reproduced here.

Rational expressions: meaning, domain and cancellation

Year 8, upper volume · text pp. 79–82 · foundation lessons 40–41 · 1/2

Name: __________ Class: ______ Date: __________

Worked example1

Simplify and retain every restriction from the original expression.

x2−9x−3\frac{x^2-9}{x-3}
Guided attempt2

Simplify and retain every restriction from the original expression.

2x2+4x2x\frac{2x^2+4x}{2x}
  1. Original restriction: x≠___.
  2. Factor the numerator as 2x(___); cancel the factor ___, not added terms.
Prerequisite3

Which displayed expression has a variable in its denominator?

1x+2\frac{1}{x+2}
  1. x²
  2. x+1
  3. 1/(x+2)
  4. x/3
Prerequisite4

Evaluate 1/(x−2) at x=3. Check the denominator first.

1/(3−2)1/(3-2)
g8-rational-expressions · 2026-10-14-upper-core-v1 · Conditions and reasons matter; measurement is not proof.

Rational expressions: meaning, domain and cancellation

Year 8, upper volume · text pp. 79–82 · foundation lessons 40–41 · 2/2

Name: __________ Class: ______ Date: __________

Independent practice5

State the domain restriction. Does the fraction simplify by cancellation?

x+1x−4\frac{x+1}{x-4}
Independent practice6

For which x is (x−5)/(x+1) equal to zero? Check that the expression is defined.

x−5x+1=0\frac{x-5}{x+1}=0
Independent practice7

Simplify and retain every restriction from the original expression.

x2−16x+4\frac{x^2-16}{x+4}
Independent practice8

Simplify and retain every restriction from the original expression.

x2+3xx\frac{x^2+3x}{x}
Exit question9

Simplify (x−2)/(x−2). Does the simplified value permit x=2? State the original restriction.

x−2x−2\frac{x-2}{x-2}
Exit question10

Simplify and retain every restriction from the original expression.

x+2x\frac{x+2}{x}
g8-rational-expressions · 2026-10-14-upper-core-v1 · Conditions and reasons matter; measurement is not proof.

Rational expressions: meaning, domain and cancellation — Further practice

Year 8, upper volume · text pp. 79–82 · foundation lessons 40–41

Further practice11

Simplify and retain every restriction from the original expression.

3x6x\frac{3x}{6x}
Further practice12

Simplify and retain every restriction from the original expression.

x2−1x−1\frac{x^2-1}{x-1}
Further practice13

Simplify and retain every restriction from the original expression.

x2+2xx+2\frac{x^2+2x}{x+2}
Further practice14

Simplify and retain every restriction from the original expression.

x−1x+1\frac{x-1}{x+1}
Extension15

After cancellation the formula is x+2. Explain why it is not the same original expression on ALL real inputs.

x2−4x−2\frac{x^2-4}{x-2}
Extension16

Simplify, and explain why x=1 remains forbidden even though the final formula is just x.

x2−xx−1\frac{x^2-x}{x-1}

Rational expressions: meaning, domain and cancellation — Teacher answers

Year 8, upper volume · text pp. 79–82 · foundation lessons 40–41

Use the worked model and guided attempt selectively, then the eight core questions. Keep further practice and extensions optional. Ask for a reason and an original-equation or expansion check.

Worked example1

Simplify and retain every restriction from the original expression.

x2−9x−3\frac{x^2-9}{x-3}
Reference answer

x+3; x≠3.

x+3x+3
  1. First record x≠3 from the original denominators and any divisor.
  2. Factor first, then cancel common factors; never cancel separate added terms.
  3. (x²-9)/(x-3) = ((x-3)(x+3))/(x-3).
  4. Therefore x+3, with x≠3 still required.
  • Use factor cancellation or a correct common denominator.
  • Retain all original restrictions, including a nonzero divisor.
Guided attempt2

Simplify and retain every restriction from the original expression.

2x2+4x2x\frac{2x^2+4x}{2x}
  1. Original restriction: x≠___.
  2. Factor the numerator as 2x(___); cancel the factor ___, not added terms.
Reference answer

x+2; x≠0.

x+2x+2
  1. First record x≠0 from the original denominators and any divisor.
  2. Factor first, then cancel common factors; never cancel separate added terms.
  3. (2x²+4x)/(2x) = (2x(x+2))/(2x).
  4. Therefore x+2, with x≠0 still required.
  • Use factor cancellation or a correct common denominator.
  • Retain all original restrictions, including a nonzero divisor.
Prerequisite3

Which displayed expression has a variable in its denominator?

1x+2\frac{1}{x+2}
  1. x²
  2. x+1
  3. 1/(x+2)
  4. x/3
Reference answer

1/(x+2)

1x+2\frac{1}{x+2}
  1. Inspect the denominator of the original expression.
  2. Only 1/(x+2) has a variable denominator; it excludes x=−2.
  • Choose the statement justified by the stated conditions and explain why.
Prerequisite4

Evaluate 1/(x−2) at x=3. Check the denominator first.

1/(3−2)1/(3-2)
Reference answer

1.

  1. The denominator is 3−2=1, which is nonzero.
  2. The value is 1/1=1.
  • Use the stated relationship and show the calculation.
  • State the result with its unit where appropriate.
Independent practice5

State the domain restriction. Does the fraction simplify by cancellation?

x+1x−4\frac{x+1}{x-4}
Reference answer

(x+1)/(x-4); x≠4.

x+1x−4\frac{x+1}{x-4}
  1. First record x≠4 from the original denominators and any divisor.
  2. No common factor cancels; solve x−4=0 to find the excluded value.
  3. (x+1)/(x-4) = (x+1)/(x-4).
  4. Therefore (x+1)/(x-4), with x≠4 still required.
  • Use factor cancellation or a correct common denominator.
  • Retain all original restrictions, including a nonzero divisor.
Independent practice6

For which x is (x−5)/(x+1) equal to zero? Check that the expression is defined.

x−5x+1=0\frac{x-5}{x+1}=0
Reference answer

x=5.

  1. A fraction is zero when the numerator is zero and the denominator is nonzero.
  2. x−5=0 gives x=5. At x=5, the denominator is 6≠0.
  3. The original fraction is 0/6=0, so x=5 is allowed.
  • Show a complete equation after clearing denominators.
  • Check original denominators and the original equality.
Independent practice7

Simplify and retain every restriction from the original expression.

x2−16x+4\frac{x^2-16}{x+4}
Reference answer

x-4; x≠-4.

x−4x-4
  1. First record x≠-4 from the original denominators and any divisor.
  2. Factor first, then cancel common factors; never cancel separate added terms.
  3. (x²-16)/(x+4) = ((x-4)(x+4))/(x+4).
  4. Therefore x-4, with x≠-4 still required.
  • Use factor cancellation or a correct common denominator.
  • Retain all original restrictions, including a nonzero divisor.
Independent practice8

Simplify and retain every restriction from the original expression.

x2+3xx\frac{x^2+3x}{x}
Reference answer

x+3; x≠0.

x+3x+3
  1. First record x≠0 from the original denominators and any divisor.
  2. Factor first, then cancel common factors; never cancel separate added terms.
  3. (x²+3x)/(x) = (x(x+3))/(x).
  4. Therefore x+3, with x≠0 still required.
  • Use factor cancellation or a correct common denominator.
  • Retain all original restrictions, including a nonzero divisor.
Exit question9

Simplify (x−2)/(x−2). Does the simplified value permit x=2? State the original restriction.

x−2x−2\frac{x-2}{x-2}
Reference answer

1; x≠2.

11
  1. First record x≠2 from the original denominators and any divisor.
  2. The common factor cancels only when it is nonzero; at x=2 the original expression is 0/0.
  3. (x-2)/(x-2) = 1.
  4. Therefore 1, with x≠2 still required.
  • Use factor cancellation or a correct common denominator.
  • Retain all original restrictions, including a nonzero divisor.
Exit question10

Simplify and retain every restriction from the original expression.

x+2x\frac{x+2}{x}
Reference answer

1+(2)/(x); x≠0.

1+2x1+\frac{2}{x}
  1. First record x≠0 from the original denominators and any divisor.
  2. Split the division over numerator terms; x is not a factor of the entire numerator x+2.
  3. (x+2)/(x) = (x)/(x)+(2)/(x).
  4. Therefore 1+(2)/(x), with x≠0 still required.
  • Use factor cancellation or a correct common denominator.
  • Retain all original restrictions, including a nonzero divisor.
Further practice11

Simplify and retain every restriction from the original expression.

3x6x\frac{3x}{6x}
Reference answer

(1)/(2); x≠0.

12\frac{1}{2}
  1. First record x≠0 from the original denominators and any divisor.
  2. Factor first, then cancel common factors; never cancel separate added terms.
  3. (3x)/(6x) = (3)/(6).
  4. Therefore (1)/(2), with x≠0 still required.
  • Use factor cancellation or a correct common denominator.
  • Retain all original restrictions, including a nonzero divisor.
Further practice12

Simplify and retain every restriction from the original expression.

x2−1x−1\frac{x^2-1}{x-1}
Reference answer

x+1; x≠1.

x+1x+1
  1. First record x≠1 from the original denominators and any divisor.
  2. Factor first, then cancel common factors; never cancel separate added terms.
  3. (x²-1)/(x-1) = ((x-1)(x+1))/(x-1).
  4. Therefore x+1, with x≠1 still required.
  • Use factor cancellation or a correct common denominator.
  • Retain all original restrictions, including a nonzero divisor.
Further practice13

Simplify and retain every restriction from the original expression.

x2+2xx+2\frac{x^2+2x}{x+2}
Reference answer

x; x≠-2.

xx
  1. First record x≠-2 from the original denominators and any divisor.
  2. Factor first, then cancel common factors; never cancel separate added terms.
  3. (x²+2x)/(x+2) = (x(x+2))/(x+2).
  4. Therefore x, with x≠-2 still required.
  • Use factor cancellation or a correct common denominator.
  • Retain all original restrictions, including a nonzero divisor.
Further practice14

Simplify and retain every restriction from the original expression.

x−1x+1\frac{x-1}{x+1}
Reference answer

(x-1)/(x+1); x≠-1.

x−1x+1\frac{x-1}{x+1}
  1. First record x≠-1 from the original denominators and any divisor.
  2. The numerator and denominator have no common nonconstant factor. Keep x≠−1.
  3. (x-1)/(x+1) = (x-1)/(x+1).
  4. Therefore (x-1)/(x+1), with x≠-1 still required.
  • Use factor cancellation or a correct common denominator.
  • Retain all original restrictions, including a nonzero divisor.
Extension15

After cancellation the formula is x+2. Explain why it is not the same original expression on ALL real inputs.

x2−4x−2\frac{x^2-4}{x-2}
Reference answer

They agree for x≠2. The original is undefined at 2, while the unrestricted polynomial has value 4.

x+2x+2
  1. First record x≠2 from the original denominators and any divisor.
  2. Factor first, then cancel common factors; never cancel separate added terms.
  3. (x²-4)/(x-2) = ((x-2)(x+2))/(x-2).
  4. Therefore x+2, with x≠2 still required.
  • Use factor cancellation or a correct common denominator.
  • Retain all original restrictions, including a nonzero divisor.
Extension16

Simplify, and explain why x=1 remains forbidden even though the final formula is just x.

x2−xx−1\frac{x^2-x}{x-1}
Reference answer

x; x≠1.

xx
  1. First record x≠1 from the original denominators and any divisor.
  2. Factor first, then cancel common factors; never cancel separate added terms.
  3. (x²-x)/(x-1) = (x(x-1))/(x-1).
  4. Therefore x, with x≠1 still required.
  • Use factor cancellation or a correct common denominator.
  • Retain all original restrictions, including a nonzero divisor.